# Finding $\int_0^\infty\frac{\sin^{2}x}{1+x^4}dx$

I am trying to evaluate $$\int_0^\infty\dfrac{\sin^{2}x}{1+x^4}dx$$ and I am stuck on how to start.

I am thinking the first step would be to substitute $$\dfrac{(1-e^{2ix})+(1-e^{-2ix})}{4}$$ for $\sin^{2}x$.

I think the singularities are at multiples of $\frac{\pi}{4}$ but I am not sure what contour to use or how to proceed from the initial substitution for $\sin^{2}x$.

Any help would be appreciated.

Thanks

We have $\sin^2 x = \frac{1-\cos(2x)}{2}$ hence: $$I = \frac{1}{4}\int_{\mathbb{R}}\frac{dx}{1+x^4}-\frac{1}{4}\,\text{Re}\left(\int_{\mathbb{R}}\frac{e^{2ix}}{1+x^4}\,dx\right)$$ and the two integrals now appearing can be computed with the residue theorem; we just need the residues in the zeroes of $1+x^4$ in the upper-half plane. By computing them we get:
$$I = \frac{\pi}{4\sqrt{2}}+\frac{\pi}{4\sqrt{2}}\left(\cos\sqrt{2}+\sin\sqrt{2}\right)e^{-\sqrt{2}}.$$
That's a good start. You also might want to notice a symmetry that lets you take the integral from $-\infty$ to $+\infty$. Now you put the "residue-calculus" tag on the question, so presumably you're going to end up looking at residues for an analytic function inside a closed contour. It's reasonable to look at a contour going from $-R$ to $+R$ on the real axis, and then back to $-R$ along a semicircle in the upper half plane. That's good for $\exp(2iz)$ but not for $\exp(-2iz)$ (why?). But again, a symmetry will come to your rescue...